arithmetically challenged people

It get's really silly now when both switch doors! :-) However neither gets

100% chance of winning and one still loses. They both can't win 66% of the time however. The problem is that the host can now only open one door if he is not to pick one already selected, and since his door now has a 1/3 chance of winning (rather than none as in the original game where he must always select one that has a goat) NO new information is provided if his is actually a goat/donkey. Therfore both contestentants now have a 50% chance of winning whether they switch or not. This in fact seems far more logical and thus obvious IMO.

Trevor.

Reply to
Trevor
Loading thread data ...

In this scenario, once the third door is opened, they each have a

50/50 chance of winning, and there is no advantage in swapping choices. Before the revelation their chances of winning were 1/3. If it seems illogical that the odds are changed by this revelation, remember that one time in three the host will reveal, not a goat, but the car. d
Reply to
Don Pearce

Only if the host opens a door at random, which isn't the case in the classic Monty Hall problem. The host knows where the car is, and always opens one of the other doors.

--
Tciao for Now!

John.
Reply to
John Williamson

Er, the host can't open one of the other doors now, since they have both already been selected by the two contestants. If he does however, and reveals a goat, then that contestant now has a ZERO chance of winning obviously, with the remaining contestant on 66%. Try actually re-reading the new scenario presented.

Trevor.

Reply to
Trevor

I think maybe you need to re-read. Each contestant picks a door, then the host opens the remaining door. If he exposes a goat, then at least one of the contestants gets a car. In fact either of the contestants will get the car, with a 50/50 chance.

d
Reply to
Don Pearce

This is called the Monty Hall problem. And yes, switching does change the probabilities of winning, and this can be proved with a simple computer program. I use this in both my math classes when we cover probabilities, and in my programming classes.

--
I'm never going to grow up.
Reply to
PeterD

NOPE, IF the host picks the remaining door he now has a 1/3 chance of the big prize, and the other two also have only a 1/3 chance each of the big prize, NOT 50:50.

trevor.

Reply to
Trevor

I should have added that yes IF (and only IF) the host reveals a goat, the other two will now have a 50:50 chance, but surely that is obvious, and remains so whether they both switch or both stay, so is NO Longer like the original MH problem at all.

Trevor.

Reply to
Trevor

The host doesn't pick the remaining door, he opens it. He reveals a goat/donkey whatever. That means the two contestants have door each, one of which has a car behind it. They now each have 50/50 odds. There is nothing in this scenario that puts one contestant's odds higher than the other since they both picked a door each at the start.

d
Reply to
Don Pearce

Of course - and that is precisely the new scenario presented. As I said - re-read. And no, it is nothing like the original MH problem.

d
Reply to
Don Pearce

ONLY if it is not behind the one the host already opened. Since he no longer has a choice of doors, he must have a 1/3 chance of showing the car.

Right, where did I say otherwise, IF the host has not already shown the car? The whole point is that the game is now no longer like the Monty Hall scenario in any way.

Trevor.

Reply to
Trevor

Why do I need to re-read, that is exactly what I said!!!!!!!!!!!!!!!!! What the hell are you objecting to??????????????????

Trevor.

Reply to
Trevor

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.